A doubt in the proof of Burnside's formula

Question

I read a proof of Burnside's formula and I got stuck in a step.

Burnside's formula

Let $X$ be a finite set, and $G$ be a finite group acting on $X$.
Denote the $G$-orbits of $X$ as $X_1,\dots,X_k$, and
$\operatorname{Fix}_X(g) = \{x \in X \mid g \cdot x = x\}$.
Then $$\bbox[yellow,5px,border:1px solid red]{k=\frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_X(g)\rvert.}$$

Proof in the link: Since $\operatorname{Fix}_X(g) = \bigcup\limits_{i=1}^k \operatorname{Fix}_{X_i}(g)$, one has $\lvert\operatorname{Fix}_X(g)\rvert = \sum\limits_{i=1}^k \lvert\operatorname{Fix}_{X_i}(g) \rvert$. \begin{align} k &= \sum\limits_{i=1}^k 1 \\ & \stackrel{?}= \sum\limits_{i=1}^k \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_{X_i}(g) \rvert \tag{?}\label{?} \\ &= \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \sum\limits_{i=1}^k \lvert\operatorname{Fix}_{X_i}(g) \rvert \\ &= \frac{1}{\lvert G \rvert} \sum\limits_{g \in G} \lvert\operatorname{Fix}_{X}(g) \rvert \end{align}

I don't understand step \eqref{?}. For each fixed $G$-orbit $X_i$, why is $\sum\limits_{g \in G} \lvert \operatorname{Fix}_{X_i}(g) \rvert = \lvert G \rvert$?

Thanks!

Answer 1

I found that I should no long left this question unanswered since it's answered in the comments. Therefore, I'll post it as community wiki so that it no longer appears in the unanswered queue.

1. In the linked text, the case for $G$ acting transitively on $X$ is first proved, so that $$ \bbox[5px, border:1px solid black]{1=|\operatorname{Orb}_X(G)|=\frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_X(g).} \tag1 \label1$$
2. Apply \eqref{1} to all $G$-orbits $X_i$. $$ 1=|\operatorname{Orb}_{X_i}(G)|=\frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_{X_i}(g) \tag2 \label2$$
3. Sum \eqref{2} over $i = 1,\dots,k$ on both sides. $$ k=\sum_{i=1}^k |\operatorname{Orb}_{X_i}(G)|=\sum_{i=1}^k \frac{1}{|G|}\sum_{g\in G}\operatorname{Fix}_{X_i}(g) \tag3 \label3$$